The latter proves that \(d{\mathbb {E}}Z \mathbf {1}(dZ>n)=O(1)\) . Indeed, suppose the contrary: for some \(n_k\rightarrow \infty \)  and \(d_k=d(n_k)\) ,
\(d_k{\mathbb {E}}Z \mathbf {1}(d_kZ> n_k)\rightarrow \infty \) . If so, the right-hand side of (REF ) goes to infinity in probability. But this is impossible, as by Theorem 1 in [1]}, \({\mathbb {P}}( d(\bar{{\mathbb {E}}}Z \mathbf {1}(dZ>n )- {\mathbb {E}}Z\mathbf {1}(d Z>n )))\leqslant (d/n) {\mathbb {E}}Z\mathbf {1}(d Z>n ) )\geqslant 1/13\)  and \( {\mathbb {E}}Z\mathbf {1}(d Z>n )\leqslant {\mathbb {E}}Z=1\) .
